2024 AMC 10B Problems/Problem 25

Solution 1

The $3x3x3$ block has side lengths of $3a, 3b, 3c$ . The $2x2x7$ block has side lengths of $2b, 2c, 7a$ .

We can create the following system of equations, knowing that the new block has 1 unit taller, deeper, and wider than the original: $3a+1 = 2b$ $3b+1=2c$ $3c+1=7a$

Adding all the equations together, we get $b+c+3 = 4a$ . Adding $a-3$ to both sides, we get $a+b+c = 4a-3$ . The question states that $a,b,c$ are all relatively prime positive integers. Therefore, our answer must be congruent to $2 \pmod{5}$ . The only answer choice satisfying this is $\boxed{92}$ . ~lprado

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2024 AMC 10B Problems/Problem 25

Solution 1